Kuhn-Tucker method

asd

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Aug 18, 2010
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How I can maximize this problem using the KKT method?

f(u,z,l)=(xz)2+l2\displaystyle f(u,z,l)=\sqrt[ ]{(xz)^2+l^2}

with the next conditions:

x3+z2+l=24\displaystyle \displaystyle\frac{x}{3}+\displaystyle\frac{z}{2}+l=24
x0\displaystyle x\geq{0};z0\displaystyle z\geq{0};l>o\displaystyle l>o
x+z>0\displaystyle x+z>0

I do the next lagrangian:

L=(xz)2l2+a(x3+z2+l24)+b(xz)\displaystyle L=-(xz)^2-l^2+a(\displaystyle\frac{x}{3}+\displaystyle\frac{z}{2}+l-24)+b(-x-z)

and then I have the KKT conditions:

2xz2+a3b0\displaystyle -2xz^2+\displaystyle\frac{a}{3}-b\geq{0} (2xz2+a3b)x=0\displaystyle (-2xz^2+\displaystyle\frac{a}{3}-b)x=0
2x2z+a2b0\displaystyle -2x^2z+\displaystyle\frac{a}{2}-b\geq{0} (2x2z+a2b)z=0\displaystyle (-2x^2z+\displaystyle\frac{a}{2}-b)z=0
2l+a0\displaystyle -2l+a\leq{0} (2l+a)l=0\displaystyle (-2l+a)l=0
x3+z2+l24=0\displaystyle \displaystyle\frac{x}{3}+\displaystyle\frac{z}{2}+l-24=0 (x3+z2+l24)a=0\displaystyle (\displaystyle\frac{x}{3}+\displaystyle\frac{z}{2}+l-24)a=0
xz0\displaystyle -x-z\leq{0} (xz)b=0\displaystyle (-x-z)b=0

But then I don't know how to get the solutions.

Thanks
 
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